SELECT T.*,
'个人客户' ERROR_TWO_TYPE
FROM (SELECT *
FROM SRBZ_HIVE.TMP_YWJH_ZQZRR_GR_02 A
WHERE EXISTS (SELECT *
FROM (SELECT CERTI_NUMBER,
COUNT(DISTINCT B.CUST_NAME)
FROM IWC_DWD.DWD_CRM_CUST_DAY B
WHERE B.STATUS_CD = '1000'
AND B.P_DAY_ID = '${date_no}'
GROUP BY B.CERTI_NUMBER
HAVING COUNT(DISTINCT B.CUST_NAME) > 1) C
WHERE C.CERTI_NUMBER = A.CERTI_NUMBER)) T
WHERE NOT EXISTS (SELECT *
FROM SRBZ_HIVE.C_IDCARD_VERIFY C
WHERE T.CERTI_NUMBER = C.IDCARD_NO
AND T.CUST_NAME = C.IDCARD_NAME)
AND NOT EXISTS
(SELECT *
FROM SRBZ_HIVE.IDENTIFY_CARD_INFO TT
WHERE T.CERTI_NUMBER = TT.IDENTIFY_CARD_ID
AND T.CUST_NAME = TT.NAME)
AND (T.REAL_FLAG <> 'T' OR T.REAL_FLAG IS NULL)
![](https://profile-avatar.csdnimg.cn/003c066a6e3e4b22946f12d5a2335662_weixin_44095378.jpg!4)
hive脚本把exists或not exist改成inner join或left join
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1条回答 默认 最新
- bj_0163_bj 2022-04-29 20:40关注
SELECT T.*, '个人客户' ERROR_TWO_TYPE FROM (SELECT * FROM SRBZ_HIVE.TMP_YWJH_ZQZRR_GR_02 A inner join (SELECT CERTI_NUMBER, COUNT(DISTINCT B.CUST_NAME) FROM IWC_DWD.DWD_CRM_CUST_DAY B WHERE B.STATUS_CD = '1000' AND B.P_DAY_ID = '${date_no}' GROUP BY B.CERTI_NUMBER HAVING COUNT(DISTINCT B.CUST_NAME) > 1) C on C.CERTI_NUMBER = A.CERTI_NUMBER) T LEFT JOIN SRBZ_HIVE.C_IDCARD_VERIFY C1 ON T.CERTI_NUMBER = C.IDCARD_NO AND T.CUST_NAME = C.IDCARD_NAME LEFT JOIN SRBZ_HIVE.IDENTIFY_CARD_INFO TT ON T.CERTI_NUMBER = TT.IDENTIFY_CARD_ID AND T.CUST_NAME = TT.NAME WHERE C1.IDCARD_NO IS NULL AND TT.IDENTIFY_CARD_ID IS NULL AND (T.REAL_FLAG <> 'T' OR T.REAL_FLAG IS NULL)
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